Richard Hertz <hertz778@gmail.com> wrote:
> On Monday, January 17, 2022 at 7:46:10 PM UTC-3, Richard Hertz wrote:
>> On Monday, January 17, 2022 at 7:27:47 PM UTC-3, bodk...@gmail.com wrote:
>
> <snip>
>>> You are discrediting yourself. You can’t even get the generic
>>> expression for the Lagrangian right.
> <snip>
>
>> Write it, alleged "mathematician", and be sure to not make a single mistake.
>>
>> I'll destroy your fucking "interpretation", woodworker. This is an issue
>> for physicists or engineers, not wood-mathematicians.
>>
>> Write it concisely and without any gobbledygook. I give you three
>> paragraphs, because I'm generous.
>>
>> But I'm going to expose you as the fucking pretender you've been since
>> you wrote your first post here. Come on, do it. And, as a BONUS
>> to mock at me with your Superior Knowledge, write the general expression
>> of the Hamiltonian. If you can, tell your audience their
>> differences, clown.
>>
>> And I'm talking about ORIGINALS, as were used 140 years ago. Not any
>> crappy post-QM interpretation. Write them EXACTLY as
>> they were conceived, for Newton's mechanics. The forum is your audience now, pretender.
>>
>> Paudkin — Maker of fine gobbledygook on whatever.
>
> Curious!
>
> Instantaneous responses for a charlatan and his small talk. Eternal
> silence for an alleged "mathematician" to reply a simple challenge
> (as Bodkin called this topic: elementary).
>
>
>

Some of us, kind sir, have a life. Are you three years old? Are you
pitching a fit in the candy aisle at the grocery store?

--
Odd Bodkin -- maker of fine toys, tools, tables

Richard Hertz <hertz778@gmail.com> wrote:
> On Monday, January 17, 2022 at 9:15:15 PM UTC-3, Ross A. Finlayson wrote:
>
> <snip>
>
>> You mean d/dt dV - dT = 0 ?
>
> No, I mean that given E = T - V = 1/2 ∑ᴺ mᵢ vᵢ² - V(r⃗₁, r⃗₂, r⃗₃, r⃗₄,
> ..., r⃗ᶰ) , for conservative systems like in newtonian gravity.

E = T + U, not T - U.

Perhaps you are confusing the Lagrangian with the total energy.

>
> This implies to solve a Lagrangian expressed as:
>
> ∂L/∂r⃗ᵢ - d/dt ∂L/∂ṙ⃗ᵢ + ∑ᴱ λᵦ ∂fᵦ/∂r⃗ᵤ = 0 , where i: 1, 2, 3,
> ...., N particles and u: 1, 2, 3, ..., E constraints.

Conservation of energy implies no such thing. The principle of stationary
action implies the Euler-Lagrange equations you just learned.

>
> ∂r⃗ᵢ ≡ (∂/∂xᵢ , ∂/∂yᵢ , ∂/∂zᵢ). Use the same notation for ∂ṙ⃗ᵢ .
>
> At any case, ask Bodkin, who knows this shit in depth.

Well, I know it more than you do. But if you scramble fast enough, you
might learn something about it in a hurry.

>
> Also, ask him how the first order Lagrangian derives in this simple
> expression used for planetary motion (even Einstein used it):
>
> 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E
>
> or
>
> d²r/dt² + μ/r² = 0 ------> 1/2 v² - μ/r = constant = E (all derived
> from Newton-Euler m d²r⃗/dt² = - GMmr⃗/r³).
>
>
>

On Tuesday, January 18, 2022 at 12:31:45 AM UTC-3, Richard Hertz wrote:

<snip>

> > You mean d/dt dV - dT = 0 ?
> No, I mean that given E = T - V = 1/2 ∑ᴺ mᵢ vᵢ² - V(r⃗₁, r⃗₂, r⃗₃, r⃗₄, ..., r⃗ᶰ) , for conservative systems like in newtonian gravity.
>
> This implies to solve a Lagrangian expressed as:
>
> ∂L/∂r⃗ᵢ - d/dt ∂L/∂ṙ⃗ᵢ + ∑ᴱ λᵦ ∂fᵦ/∂r⃗ᵤ = 0 , where i: 1, 2, 3, ...., N particles and u: 1, 2, 3, ..., E constraints.
>
> ∂r⃗ᵢ ≡ (∂/∂xᵢ , ∂/∂yᵢ , ∂/∂zᵢ). Use the same notation for ∂ṙ⃗ᵢ .
>
> At any case, ask Bodkin, who knows this shit in depth.
>
> Also, ask him how the first order Lagrangian derives in this simple expression used for planetary motion (even Einstein used it):
>
> 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E
>
> or
>
> d²r/dt² + μ/r² = 0 ------> 1/2 v² - μ/r = constant = E (all derived from Newton-Euler m d²r⃗/dt² = - GMmr⃗/r³).

Paul, Bodkin, you should have known this better.

Besides the work of Euler on Newton's theories (being Euler "the one from whom we learned it all", Lagrange dixit),
the main question behind these complex developments is a simple question that has not answered in 300 years:

Is the Solar System stable, or it just seems to be so?

As was elementary to deduce, if you create a complex set of equations composed by a given number N of less complex equations
which are STABLE, the result will be poisoned by the premises embedded into each elementary set: they are stable because not a
single perturbation was introduced from the start. So, the result of accumulating a set of stable equations is STABILITY.

So Euler, knowing that, introduced SMALL PERTURBATIONS in every single set of opposing forces, to see what happened with the
big picture of N-body systems under small perturbations of their stability, in order to analyze IF in the big picture, stability settled on
or the entire relationship of forces couldn't converge, with time, to such apparent stability that was witnessed in the solar system.

Lagrange, learning from Euler, worked the perturbations from the point of view of energies (not forces, as Euler did). And the key for
both development was the introduction since the start of small perturbations that altered the apparent equilibrium.

Besides Laplace and his celestial mechanics (he coined the term), the fourth scientist in a row (in 80 years) was le Verrier.

Euler and Lagrange theories of perturbations were not concerned about orbital mechanics per se, as Laplace and le Verrier did.

At any case, to use these theories in classrooms, they have to be simplified by limiting the number N to 2 or 3, and this is what
is taught even today, as it can be seen on any book or online site dealing with planetary motion, orbital mechanics, etc. And every
conic figure (circles, ellipses, parabolas or hyperbolas) are treated with more or less interest.

The use of full-fledged theories of perturbations are left (costs, complexities) to current institutions that deal with the complexity
of several types of systems, were conservation is expected or not. Branches of these original theories incorporated relativity (in physics)
or transformed the problem in analogies of complex interactions of events (with the correspondent adaptation of forces or energies to
any other manifestation). One example is a branch of analysis of the behavior of stock markets or currencies worldwide.

But, the most widely use of these theories are for conservative systems. A highly non-linear complexity that only can be solved by
numerical simulation on computers, as no analytical solution is known when N > 3. And for N = 3, the result is based on first degree
approximations, with a large number of simplifications.

So, what Paul wrote is the simplest derivation for N = 2, which is widely taught and used. Above 2, it's highly speculative due to forced
simplifications to go ahead.

And Bodkin, stop projecting on others your mechanism for cheating here on whatever the subject is.
Other persons understand things better, either you like or (seems obvious) fucking hate it.

Take another red pill and learn to deal with this.

Richard Hertz <hertz778@gmail.com> wrote:
> On Tuesday, January 18, 2022 at 12:31:45 AM UTC-3, Richard Hertz wrote:
>
> <snip>
>
>>> You mean d/dt dV - dT = 0 ?
>> No, I mean that given E = T - V = 1/2 ∑ᴺ mᵢ vᵢ² - V(r⃗₁, r⃗₂, r⃗₃, r⃗₄,
>> ..., r⃗ᶰ) , for conservative systems like in newtonian gravity.
>>
>> This implies to solve a Lagrangian expressed as:
>>
>> ∂L/∂r⃗ᵢ - d/dt ∂L/∂ṙ⃗ᵢ + ∑ᴱ λᵦ ∂fᵦ/∂r⃗ᵤ = 0 , where i: 1, 2, 3, ...., N
>> particles and u: 1, 2, 3, ..., E constraints.
>>
>> ∂r⃗ᵢ ≡ (∂/∂xᵢ , ∂/∂yᵢ , ∂/∂zᵢ). Use the same notation for ∂ṙ⃗ᵢ .
>>
>> At any case, ask Bodkin, who knows this shit in depth.
>>
>> Also, ask him how the first order Lagrangian derives in this simple
>> expression used for planetary motion (even Einstein used it):
>>
>> 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E
>>
>> or
>>
>> d²r/dt² + μ/r² = 0 ------> 1/2 v² - μ/r = constant = E (all derived from
>> Newton-Euler m d²r⃗/dt² = - GMmr⃗/r³).
>
> Paul, Bodkin, you should have known this better.
>
> Besides the work of Euler on Newton's theories (being Euler "the one from
> whom we learned it all", Lagrange dixit),
> the main question behind these complex developments is a simple question
> that has not answered in 300 years:
>
> Is the Solar System stable, or it just seems to be so?
>
> As was elementary to deduce, if you create a complex set of equations
> composed by a given number N of less complex equations
> which are STABLE, the result will be poisoned by the premises embedded
> into each elementary set: they are stable because not a
> single perturbation was introduced from the start. So, the result of
> accumulating a set of stable equations is STABILITY.
>
> So Euler, knowing that, introduced SMALL PERTURBATIONS in every single
> set of opposing forces, to see what happened with the
> big picture of N-body systems under small perturbations of their
> stability, in order to analyze IF in the big picture, stability settled on
> or the entire relationship of forces couldn't converge, with time, to
> such apparent stability that was witnessed in the solar system.
>
> Lagrange, learning from Euler, worked the perturbations from the point of
> view of energies (not forces, as Euler did). And the key for
> both development was the introduction since the start of small
> perturbations that altered the apparent equilibrium.

No. The principle of stationary action does not have to do with stability
of trajectory from an energy minimum perspective.

I’m sorry, you’re just not getting the point of the action principle at
all. This isn’t really a surprise. It’s a topic for a 3rd or 4th year
physics student and usually engineers have diverged to their own versions
of mechanics classes by then, or for EE candidates, skipping that subject
entirely.

>
> Besides Laplace and his celestial mechanics (he coined the term), the
> fourth scientist in a row (in 80 years) was le Verrier.
>
> Euler and Lagrange theories of perturbations were not concerned about
> orbital mechanics per se, as Laplace and le Verrier did.
>
> At any case, to use these theories in classrooms, they have to be
> simplified by limiting the number N to 2 or 3, and this is what
> is taught even today, as it can be seen on any book or online site
> dealing with planetary motion, orbital mechanics, etc. And every
> conic figure (circles, ellipses, parabolas or hyperbolas) are treated
> with more or less interest.
>
> The use of full-fledged theories of perturbations are left (costs,
> complexities) to current institutions that deal with the complexity
> of several types of systems, were conservation is expected or not.
> Branches of these original theories incorporated relativity (in physics)
> or transformed the problem in analogies of complex interactions of events
> (with the correspondent adaptation of forces or energies to
> any other manifestation). One example is a branch of analysis of the
> behavior of stock markets or currencies worldwide.
>
> But, the most widely use of these theories are for conservative systems.
> A highly non-linear complexity that only can be solved by
> numerical simulation on computers, as no analytical solution is known
> when N > 3. And for N = 3, the result is based on first degree
> approximations, with a large number of simplifications.
>
> So, what Paul wrote is the simplest derivation for N = 2, which is widely
> taught and used. Above 2, it's highly speculative due to forced
> simplifications to go ahead.
>
> And Bodkin, stop projecting on others your mechanism for cheating here on
> whatever the subject is.
> Other persons understand things better, either you like or (seems obvious) fucking hate it.
>
> Take another red pill and learn to deal with this.
>
>
>
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On Tuesday, January 18, 2022 at 1:16:58 PM UTC-3, bodk...@gmail.com wrote:

<snip>

> No. The principle of stationary action does not have to do with stability
> of trajectory from an energy minimum perspective.
>
> I’m sorry, you’re just not getting the point of the action principle at
> all. This isn’t really a surprise. It’s a topic for a 3rd or 4th year
> physics student and usually engineers have diverged to their own versions
> of mechanics classes by then, or for EE candidates, skipping that subject
> entirely.

Did you read a single reference to Hamilton in my posts, Bodkin?

I told you several times here, Bodkin: FUNDAMENTALS! Always go to the fucking FUNDAMENTALS!

No wonder you never progress in your narratives. You can't find the FUNDAMENTALS on whatever, so you start backwards.

And because of that you are digressing permanently, trying to fit the facts withing any given timeline from the end to the beginning.

And that's why your thought process is flawed. You did good by focusing on woodworking. It fits the material of which your brain is made.

Don't blame me, blame nature. It's not a matter of physics vs. engineering.

It's a matter of critical thinking, analytical power and logic. Some have it a lot and some others have less of it.

You are the kind of people that explain microelectronics starting in 2021, going backwards to de Forest, in 1907.

And then, claim anyone for being ignorant or uninformed if is not using the latest to explain the original. FUNDAMENTALS, Bodkin.

Hamilton was 70 years ahead of Lagrange, when his ideas were widely accepted and Hamilton wanted to chip in with his version.

But you have to analyze the timeline Newton-Euler FIRST, if you want to get Hamilton, passing by Lagrange, Laplace, D'Alembert, etc.

What moved Euler to think about this topic? That's FUNDAMENTAL, Bodkin. The rest is almost anecdotal.

Richard Hertz <hertz778@gmail.com> wrote:
> On Tuesday, January 18, 2022 at 1:16:58 PM UTC-3, bodk...@gmail.com wrote:
>
> <snip>
>
>> No. The principle of stationary action does not have to do with stability
>> of trajectory from an energy minimum perspective.
>>
>> I’m sorry, you’re just not getting the point of the action principle at
>> all. This isn’t really a surprise. It’s a topic for a 3rd or 4th year
>> physics student and usually engineers have diverged to their own versions
>> of mechanics classes by then, or for EE candidates, skipping that subject
>> entirely.
>
> Did you read a single reference to Hamilton in my posts, Bodkin?

No. Still, the difference is that I know what Hamiltons and Lagrangians
are, and why they are useful, and you are still grappling to figure that
out.

So when you first attempted to describe Lagrangians with conjugate
variables, you had no idea what you were presenting was closer to a
Hamiltonian than to a Lagrangian. I *recognized* the Hamiltonian features
because they are *familiar* to me, because I have *studied* them. One other
key difference is that I do not open my yap about things I do not know
about, like RF antenna design, while you feel it’s beneficial to talk first
and learn later.

>
> I told you several times here, Bodkin: FUNDAMENTALS! Always go to the fucking FUNDAMENTALS!
>
> No wonder you never progress in your narratives. You can't find the
> FUNDAMENTALS on whatever, so you start backwards.
>
> And because of that you are digressing permanently, trying to fit the
> facts withing any given timeline from the end to the beginning.
>
> And that's why your thought process is flawed. You did good by focusing
> on woodworking. It fits the material of which your brain is made.
>
> Don't blame me, blame nature. It's not a matter of physics vs. engineering.
>
> It's a matter of critical thinking, analytical power and logic. Some have
> it a lot and some others have less of it.
>
> You are the kind of people that explain microelectronics starting in
> 2021, going backwards to de Forest, in 1907.
>
> And then, claim anyone for being ignorant or uninformed if is not using
> the latest to explain the original. FUNDAMENTALS, Bodkin.
>
> Hamilton was 70 years ahead of Lagrange, when his ideas were widely
> accepted and Hamilton wanted to chip in with his version.
>
> But you have to analyze the timeline Newton-Euler FIRST, if you want to
> get Hamilton, passing by Lagrange, Laplace, D'Alembert, etc.
>
> What moved Euler to think about this topic? That's FUNDAMENTAL, Bodkin.
> The rest is almost anecdotal.
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On Tuesday, January 18, 2022 at 9:25:29 AM UTC-3, Paul B. Andersen wrote:
> Den 17.01.2022 20:36, skrev Richard Hertz:

<snip>

> I suppose that L is the Lagrangian.
>
> L = T - V kinetic energy MINUS potential energy.
>
> L is constant only if T and V both are constant.
> (Example: planet in circular orbit)
>
> But in the general case, both T and V may change.
> (Example planet in orbit with eccentricity ≠ 0)
>
> Since the total energy E = T+V = constant,
> dT/dt = - dV/dt
>
> dL/dt = 2dT/dt

Mr. Astrophysicist, shame on you!

The most elementary application for a system with conserved energy is with celestial bodies orbitals.

This equation is valid for bounded (planets and their moons) or unbounded orbits (asteroids that pass by).

1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E

For bounded orbits, E < 0 and 0 <= e < 1 (excentricity of the trajectory; i.e.: e = 0.2025 for Mercury).

In the case of unbounded hyperbolic orbits, E > 0 and e > 1.

So, at infinity, where gravitational potential goes to ZERO, and angular moment vanished because θ reached a limit value,
so dθ/dt = 0, only remains KE at infinity: E = 1/2 m v²∞.

Your dT/dt = 0 at infinity because T = constant = E

but your dV/dt --> 0 just because V ---> 0 for distances very far away from the gravitational source (the Sun, in this case).

This is the fate or a one time flyby of an asteroid/comet if E > 0 (hyperbolic trajectory).

In some cases, E = 0 and e = 1. It describes a parabolic trajectory, an exceptional case.

Your dT/dt = - dV/dt applies for elliptic orbits only.

and your dL/dt = 2dT/dt IS WRONG! We said that energy is CONSERVED, as well as angular moment.

Go to the corner!

Den 18.01.2022 17:06, skrev Richard Hertz:

>
> So, what Paul wrote is the simplest derivation for N = 2, which is widely taught and used. Above 2, it's highly speculative due to forced
> simplifications to go ahead.
>

What Paul wrote was not a derivation of anything,
he only pointed out your error when YOU wrote:

> Den 17.01.2022 20:36, skrev Richard Hertz:
>> Fucking ignorant you all!
>>
>> Lagrangians, Hamiltonian and similar shit are basically Newton's laws for mechanical energy, WHICH DISIPATES with enough time (say a
couple of billion years for celestial mechanics). That's why dL/dt = 0
and its integral renders L = CONSTANT.

This is wrong because:
L = T - V kinetic energy MINUS potential energy.
L is constant only if T and V both are constant.
(Example: planet in circular orbit)
But in the general case, both T and V may change.
(Example planet in orbit with eccentricity ≠ 0)

Since the total energy E = T+V = constant,
dT/dt = - dV/dt

dL/dt = 2dT/dt

Now you are desperate to divert the attention
from the fact that you didn't know what the Lagrangian is.

You seem to have mixed up the total energy E and
Lagrangian L as exemplified when you wrote:

> Den 17.01.2022 21:56, skrev Richard Hertz:>
>> 6) The most used derivation of the lagrangian E = T + U for celestial mechanics, if for CONSERVATIVE systems, where E = constant.

But the total energy E is NOT the Lagrangian!.

> Den 17.01.2022 21:56, skrev Richard Hertz:
>> 7) The Euler-Lagrange has proven to be useful for systems that are conservative OR NOT. For the classic Lagrangian, widely used for cosmic or atomic interactions, E is conserved. So, in such systems, dE/dt = 0. This, if properly developed with additional perturbations to what is already A VERY COMPLEX SET OF EQUATIONS, is what rules Newton dynamics for spatial systems (NASA, Russia, etc.)

So why didn't you mention the The Euler-Lagrange equation?
If the coordinates of the system are (q₁, q₂, q₃, .. qᵢ)
you get i simultaneous differential equations:
d/dt(∂L/∂q̇ᵢ) = ∂L/∂qᵢ

There is no dE/dt in this "VERY COMPLEX SET OF EQUATIONS".

--
Paul

https://paulba.no/

Richard Hertz <hertz778@gmail.com> wrote:
> On Tuesday, January 18, 2022 at 9:25:29 AM UTC-3, Paul B. Andersen wrote:
>> Den 17.01.2022 20:36, skrev Richard Hertz:
>
> <snip>
>
>> I suppose that L is the Lagrangian.
>>
>> L = T - V kinetic energy MINUS potential energy.
>>
>> L is constant only if T and V both are constant.
>> (Example: planet in circular orbit)
>>
>> But in the general case, both T and V may change.
>> (Example planet in orbit with eccentricity ≠ 0)
>>
>> Since the total energy E = T+V = constant,
>> dT/dt = - dV/dt
>>
>> dL/dt = 2dT/dt
>
> Mr. Astrophysicist, shame on you!
>
> The most elementary application for a system with conserved energy is
> with celestial bodies orbitals.
>
> This equation is valid for bounded (planets and their moons) or unbounded
> orbits (asteroids that pass by).
>
> 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E

Oh, look, you want to talk about total energy.

This conversation had been about Lagrangians, but I guess you’ve discovered
you don’t know anything about them after all and are trying to divert the
conversation.

Once again, I’ll just point out that the conservation of energy applied to
celestial orbits is something that is routinely done in first and second
year physics classes, while Lagrangian mechanics are typically covered in
third and fourth year physics. So it’s no wonder you want to stick to the
things you know — that’s a GOOD thing. Just don’t try to pretend you’re
still talking about Lagrangian mechanics.

>
> For bounded orbits, E < 0 and 0 <= e < 1 (excentricity of the trajectory;
> i.e.: e = 0.2025 for Mercury).
>
> In the case of unbounded hyperbolic orbits, E > 0 and e > 1.
>
> So, at infinity, where gravitational potential goes to ZERO, and angular
> moment vanished because θ reached a limit value,
> so dθ/dt = 0, only remains KE at infinity: E = 1/2 m v²∞.
>
> Your dT/dt = 0 at infinity because T = constant = E
>
> but your dV/dt --> 0 just because V ---> 0 for distances very far away
> from the gravitational source (the Sun, in this case).
>
> This is the fate or a one time flyby of an asteroid/comet if E > 0 (hyperbolic trajectory).
>
> In some cases, E = 0 and e = 1. It describes a parabolic trajectory, an exceptional case.
>
> Your dT/dt = - dV/dt applies for elliptic orbits only.
>
> and your dL/dt = 2dT/dt IS WRONG! We said that energy is CONSERVED, as
> well as angular moment.
>
> Go to the corner!
>
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On Tuesday, January 18, 2022 at 12:31:45 AM UTC-3, Richard Hertz wrote:

<snip>

AGAIN, AND AGAIN IF NECESSARY!

Lagrange developed his theory based on the total energy of any system with N particles under Newton's laws, which differentiated
from the early work of the SupremeThinker Euler (not Bodkin), who took Newton's theories to the realm of true calculus and used
FORCES and the net remain of them in any system with N bodies. Net remain (forces): Net remain (energies), two ways to observe
and describe the same problem.

Stated in a general way, it is the TRUE focus of Lagrange. Then he followed with HIS perturbation theory, which applied for a 3-Body
system, due to which he found zones of equilibrium in the System Sun-Earth-Moon, which are currently used to position satellites.
He devoted a lot of his work to introduce perturbations in the equations, as a method to find out convergence or divergence.

The equations below are the pure essence of his basic work, with perturbations introduced as constraints.

But Paul and Bodkin are reacting as persons with deep butthurt. Why? Did you took offense of my "irruption" in your "realm"?

Learn how to write a post if you don't be molested, then, stooges.

> > You mean d/dt dV - dT = 0 ?
> No, I mean that given E = T - V = 1/2 ∑ᴺ mᵢ vᵢ² - V(r⃗₁, r⃗₂, r⃗₃, r⃗₄, ..., r⃗ᶰ) , for conservative systems like in newtonian gravity.
>
> This implies to solve a Lagrangian expressed as:
>
> ∂L/∂r⃗ᵢ - d/dt ∂L/∂ṙ⃗ᵢ + ∑ᴱ λᵦ ∂fᵦ/∂r⃗ᵤ = 0 , where i: 1, 2, 3, ...., N particles and u: 1, 2, 3, ..., E constraints.
>
> ∂r⃗ᵢ ≡ (∂/∂xᵢ , ∂/∂yᵢ , ∂/∂zᵢ). Use the same notation for ∂ṙ⃗ᵢ .
>
> At any case, ask Bodkin, who knows this shit in depth.
>
> Also, ask him how the first order Lagrangian derives in this simple expression used for planetary motion (even Einstein used it):
>
> 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E
>
> or
>
> d²r/dt² + μ/r² = 0 ------> 1/2 v² - μ/r = constant = E (all derived from Newton-Euler m d²r⃗/dt² = - GMmr⃗/r³).

https://mathshistory.st-andrews.ac.uk/Biographies/Lagrange/

...............
In 1756 Lagrange sent Euler results that he had obtained on applying the calculus of variations to mechanics. These results generalised results which Euler had himself obtained and Euler consulted Maupertuis, the president of the Berlin Academy, about this remarkable young mathematician. Not only was Lagrange an outstanding mathematician but he was also a strong advocate for the principle of least action so Maupertuis had no hesitation but to try to entice Lagrange to a position in Prussia. He arranged with Euler that he would let Lagrange know that the new position would be considerably more prestigious than the one he held in Turin. However, Lagrange did not seek greatness, he only wanted to be able to devote his time to mathematics, and so he shyly but politely refused the position.
...............
The papers by Lagrange which appear in these transactions cover a variety of topics. He published his beautiful results on the calculus of variations, and a short work on the calculus of probabilities. In a work on the foundations of dynamics, Lagrange based his development on the principle of least action and on kinetic energy.
...............
In papers which were published in the third volume, Lagrange studied the integration of differential equations and made various applications to topics such as fluid mechanics (where he introduced the Lagrangian function). Also contained are methods to solve systems of linear differential equations which used the characteristic value of a linear substitution for the first time. Another problem to which he applied his methods was the study the orbits of Jupiter and Saturn.
..............
Returning to Turin in early 1765, Lagrange entered, later that year, for the Académie des Sciences prize of 1766 on the orbits of the moons of Jupiter. D'Alembert, who had visited the Berlin Academy and was friendly with Frederick II of Prussia, arranged for Lagrange to be offered a position in the Berlin Academy. Despite no improvement in Lagrange's position in Turin, he again turned the offer down writing:-
...............
By March 1766 d'Alembert knew that Euler was returning to St Petersburg and wrote again to Lagrange to encourage him to accept a post in Berlin. Full details of the generous offer were sent to him by Frederick II in April, and Lagrange finally accepted. Leaving Turin in August, he visited d'Alembert in Paris, then Caraccioli in London before arriving in Berlin in October. Lagrange succeeded Euler as Director of Mathematics at the Berlin Academy on 6 November 1766.
...............
Turin always regretted losing Lagrange and from time to time his return there was suggested, for example in 1774. However, for 20 years Lagrange worked at Berlin, producing a steady stream of top quality papers and regularly winning the prize from the Académie des Sciences of Paris. He shared the 1772 prize on the three body problem with Euler, won the prize for 1774, another one on the motion of the moon, and he won the 1780 prize on perturbations of the orbits of comets by the planets.

His work in Berlin covered many topics: astronomy, the stability of the solar system, mechanics, dynamics, fluid mechanics, probability, and the foundations of the calculus. He also worked on number theory proving in 1770 that every positive integer is the sum of four squares. In 1771 he proved Wilson's theorem (first stated without proof by Waring) that nnn is prime if and only if (n−1)!+1(n -1)! + 1(n−1)!+1 is divisible by nnn. In 1770 he also presented his important work Réflexions sur la résolution algébrique des équations Ⓣ which made a fundamental investigation of why equations of degrees up to 4 could be solved by radicals. The paper is the first to consider the roots of an equation as abstract quantities rather than having numerical values. He studied permutations of the roots and, although he does not compose permutations in the paper, it can be considered as a first step in the development of group theory continued by Ruffini, Galois and Cauchy.

Although Lagrange had made numerous major contributions to mechanics, he had not produced a comprehensive work. He decided to write a definitive work incorporating his contributions and wrote to Laplace on 15 September 1782:-

I have almost completed a 'Traité de mécanique analytique' Ⓣ, based uniquely on the principle of virtual velocities; but, as I do not yet know when or where I shall be able to have it printed, I am not rushing to put the finishing touches to it.
...............

ENERGY AND PERTURBATION ANALYSIS WERE HIS GOALS ON NEWTONIAN MECHANICS.

SO, SHUT THE FUCK UP!

Richard Hertz <hertz778@gmail.com> wrote:
> https://mathshistory.st-andrews.ac.uk/Biographies/Lagrange/
>
> ..............
> In 1756 Lagrange sent Euler results that he had obtained on applying the
> calculus of variations

Yes, variations over WHAT?

> to mechanics. These results generalised results which Euler had himself
> obtained and Euler consulted Maupertuis, the president of the Berlin
> Academy, about this remarkable young mathematician. Not only was Lagrange
> an outstanding mathematician but he was also a strong advocate for the
> principle of least action

Indeed, and what does “least action” mean?

> so Maupertuis had no hesitation but to try to entice Lagrange to a
> position in Prussia. He arranged with Euler that he would let Lagrange
> know that the new position would be considerably more prestigious than
> the one he held in Turin. However, Lagrange did not seek greatness, he
> only wanted to be able to devote his time to mathematics, and so he shyly
> but politely refused the position.

As you say, the basics are important.
--
Odd Bodkin -- maker of fine toys, tools, tables

On Tuesday, January 18, 2022 at 6:45:11 PM UTC-3, bodk...@gmail.com wrote:

<snip>

> Indeed, and what does “least action” mean?
.......
> As you say, the basics are important.

Think, Bodkin, think! I can't do it for you!

LEAST ACTION:::LEAST EFFORT:::LEAST WASTE OF ENERGY FOR ANY GIVEN WORK!

Lazy and ignorant people is borne with such principle on their DNA!

Nature tends to do things in the most effective way, wasting the least energy possible.

Is it so difficult to represent this using the new tool of calculus (for that epoch)?

Which is the trajectory by which I (whoever or whichever) will work the least (expend the lowest amount of energy)?

Lagrange and his view of newtonian mechanics is IMPREGNATED by these FUNDAMENTALS: Energy and least action!

Hard to accept such simplicity, isn't it, relativist?

FUNDAMENTALS!

All the time.

Richard Hertz <hertz778@gmail.com> wrote:
> On Tuesday, January 18, 2022 at 6:45:11 PM UTC-3, bodk...@gmail.com wrote:
>
> <snip>
>
>> Indeed, and what does “least action” mean?
> ......
>> As you say, the basics are important.
>
> Think, Bodkin, think! I can't do it for you!
>
> LEAST ACTION:::LEAST EFFORT:::LEAST WASTE OF ENERGY FOR ANY GIVEN WORK!

Nope. Don’t guess.

Action has a specific meaning in physics, as does the principle of least
action.

Now, you can either open a book and start reading, or you can decide you
don’t care what it means and stop talking about it, or you can continue to
make shit up just because it’s what you like to do.

I know you have no dignity and don’t care what your choices say about you.
You have deep disdain for people in general, and I suppose for yourself as
well, so dignity is not really a concern for you.

Since none of your engineering texts apparently tell you about the
principle of stationary action, do you need a recommendation for a physics
book? I have several I can recommend.

>
> Lazy and ignorant people is borne with such principle on their DNA!
>
> Nature tends to do things in the most effective way, wasting the least energy possible.
>
> Is it so difficult to represent this using the new tool of calculus (for that epoch)?
>
> Which is the trajectory by which I (whoever or whichever) will work the
> least (expend the lowest amount of energy)?
>
> Lagrange and his view of newtonian mechanics is IMPREGNATED by these
> FUNDAMENTALS: Energy and least action!
>
> Hard to accept such simplicity, isn't it, relativist?
>
> FUNDAMENTALS!
>
> All the time.
>
>

--
Odd Bodkin -- maker of fine toys, tools, tables

Den 18.01.2022 21:41, skrev Richard Hertz:
> On Tuesday, January 18, 2022 at 9:25:29 AM UTC-3, Paul B. Andersen wrote:
>> Den 17.01.2022 20:36, skrev Richard Hertz:
<unsnip>>>> Fucking ignorant you all!
>>>
>>> Lagrangians, Hamiltonian and similar shit are basically Newton's laws for mechanical energy, WHICH DISIPATES with enough time (say a
>>> couple of billion years for celestial mechanics).
>>>
>>> That's why dL/dt = 0 and its integral renders L = CONSTANT.

I can understand why you snipped your own wrong statement! :-D

>> I suppose that L is the Lagrangian.
>>
>> L = T - V kinetic energy MINUS potential energy.
>>
>> L is constant only if T and V both are constant.
>> (Example: planet in circular orbit)
>>
>> But in the general case, both T and V may change.
>> (Example planet in orbit with eccentricity ≠ 0)
>>
>> Since the total energy E = T+V = constant,
>> dT/dt = - dV/dt
>>
>> dL/dt = 2dT/dt

Is the following supposed to be a defence for your claim that dL/dt = 0?

>
> Mr. Astrophysicist, shame on you!
>
> The most elementary application for a system with conserved energy is with celestial bodies orbitals.
>
> This equation is valid for bounded (planets and their moons) or unbounded orbits (asteroids that pass by).
>
> 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E

Yes, but this is the total energy, not the Lagrangian.

T = 1/2 m [r² (dθ/dt)² + (dr/dt)²]
V = – GMm/r

L = T-V = 1/2 m [r² (dθ/dt)² + (dr/dt)²] + GMm/r

E is constant, L is NOT.

The next step is to use the Euler-Lagrange equation
which will in this case give two equations:

d/dt(∂L/∂ṙ) = ∂L/∂r

d/dt(∂L/∂θ̇ ) = ∂L/∂θ

>
> For bounded orbits, E < 0 and 0 <= e < 1 (excentricity of the trajectory; i.e.: e = 0.2025 for Mercury).
>
> In the case of unbounded hyperbolic orbits, E > 0 and e > 1.
>
> So, at infinity, where gravitational potential goes to ZERO, and angular moment vanished because θ reached a limit value,
> so dθ/dt = 0, only remains KE at infinity: E = 1/2 m v²∞.

But still: T + V = E = const

lim T = const when r -> ∞
lim V = 0 when r -> ∞
lim E = const when r -> ∞ as always

>
> Your dT/dt = 0 at infinity because T = constant = E
>
> but your dV/dt --> 0 just because V ---> 0 for distances very far away from the gravitational source (the Sun, in this case).

So?

dV/dt -> 0 and dT/dt -> 0 when r -> ∞

What was your point?

>
> This is the fate or a one time flyby of an asteroid/comet if E > 0 (hyperbolic trajectory).

Quite. But T+V = E is still constant

>
> In some cases, E = 0 and e = 1. It describes a parabolic trajectory, an exceptional case.

Quite. But T+V = E is still constant

>
> Your dT/dt = - dV/dt applies for elliptic orbits only.

No, it follows trivially from E = T + V. Always.

dT/dt + dv/dt = dE/dt = 0
dT/dt = -dV/dt

>
> and your dL/dt = 2dT/dt IS WRONG! We said that energy is CONSERVED, as well as angular moment.

L = T - V
dL/dt = dT/dt - dV/dt = 2dT/dt

BECAUSE total energy is CONSERVED

>
> Go to the corner!
>

No, I will leave you there alone.

--
Paul

https://paulba.no/

On Tuesday, January 18, 2022 at 7:21:38 PM UTC-3, bodk...@gmail.com wrote:

<snip>

> >> Indeed, and what does “least action” mean?
> >> As you say, the basics are important.
> >
> > Think, Bodkin, think! I can't do it for you!
> >
> > LEAST ACTION:::LEAST EFFORT:::LEAST WASTE OF ENERGY FOR ANY GIVEN WORK!
> Nope. Don’t guess.
>
> Action has a specific meaning in physics, as does the principle of least action.
............
> Since none of your engineering texts apparently tell you about the
> principle of stationary action, do you need a recommendation for a physics
> book? I have several I can recommend.
> >
> > Lazy and ignorant people is borne with such principle on their DNA!
> > Nature tends to do things in the most effective way, wasting the least energy possible.
> >
> > Is it so difficult to represent this using the new tool of calculus (for that epoch)?
> >
> > Which is the trajectory by which I (whoever or whichever) will work the least (expend the lowest amount of energy)?
> >
> > Lagrange and his view of newtonian mechanics is IMPREGNATED by these FUNDAMENTALS: Energy and least action!
> >
> > Hard to accept such simplicity, isn't it, relativist?
> >
> > FUNDAMENTALS!

What is the principle of least action in classical mechanics?
In Classical Mechanics when a particle moves from one initial point to a final point the path that it will follow is the one
where action is minimum.
------------
The stationary-action principle – also known as the principle of least action – is a variational principle that, when applied to
the action of a mechanical system, yields the equations of motion for that system. ...
-----------
What is classical mechanics action?
More formally, action is a mathematical functional which takes the trajectory, also called path or history, of the system as its
argument and has a real number as its result. ... Generally, the action takes different values for different paths.
-----------
https://www.feynmanlectures.caltech.edu/II_19.html

The Principle of Least Action

“Mr. Bader told me the following: Suppose you have a particle (in a gravitational field, for instance) which starts somewhere and
moves to some other point by free motion—you throw it, and it goes up and comes down (Fig. 19–1). It goes from the original
place to the final place in a certain amount of time. Now, you try a different motion. Suppose that to get from here to there, it went
as shown in Fig. 19–2 but got there in just the same amount of time.. Then he said this: If you calculate the kinetic energy at every
moment on the path, take away the potential energy, and integrate it over the time during the whole path, you’ll find that the number
you’ll get is bigger than that for the actual motion.

“In other words, the laws of Newton could be stated not in the form F=ma but in the form: the average kinetic energy less the average potential energy is as little as possible for the path of an object going from one point to another.

“Let me illustrate a little bit better what it means. If you take the case of the gravitational field, then if the particle has the path x(t) (let’s just take one dimension for a moment; we take a trajectory that goes up and down and not sideways), where x is the height above the ground, the kinetic energy is 1/2m(dx/dt)², and the potential energy at any time is mgx. Now I take the kinetic energy minus the potential energy at every moment along the path and integrate that with respect to time from the initial time to the final time. Let’s suppose that at the original time t1 we started at some height and at the end of the time t2 we are definitely ending at some other place (Fig. 19–3).

“Then the integral is ∫ [1/2m(dx/dt)²−mgx] dt, between t1 and t2

The actual motion is some kind of a curve—it’s a parabola if we plot against the time—and gives a certain value for the integral. But we could imagine some other motion that went very high and came up and down in some peculiar way (Fig. 19–4). We can calculate the kinetic energy minus the potential energy and integrate for such a path … or for any other path we want.

The miracle is that the true path is the one for which that integral is least.
---------

Enough? I can find more trusted sources about the ORIGINAL PRINCIPLE OF LEAST ACTION (not the relativistic shit, staining everything!).

Richard Hertz <hertz778@gmail.com> wrote:
> On Tuesday, January 18, 2022 at 7:21:38 PM UTC-3, bodk...@gmail.com wrote:
>
> <snip>
>
>>>> Indeed, and what does “least action” mean?
>>>> As you say, the basics are important.
>>>
>>> Think, Bodkin, think! I can't do it for you!
>>>
>>> LEAST ACTION:::LEAST EFFORT:::LEAST WASTE OF ENERGY FOR ANY GIVEN WORK!
>> Nope. Don’t guess.
>>
>> Action has a specific meaning in physics, as does the principle of least action.
> ...........
>> Since none of your engineering texts apparently tell you about the
>> principle of stationary action, do you need a recommendation for a physics
>> book? I have several I can recommend.
>>>
>>> Lazy and ignorant people is borne with such principle on their DNA!
>>> Nature tends to do things in the most effective way, wasting the least energy possible.
>>>
>>> Is it so difficult to represent this using the new tool of calculus (for that epoch)?
>>>
>>> Which is the trajectory by which I (whoever or whichever) will work the
>>> least (expend the lowest amount of energy)?
>>>
>>> Lagrange and his view of newtonian mechanics is IMPREGNATED by these
>>> FUNDAMENTALS: Energy and least action!
>>>
>>> Hard to accept such simplicity, isn't it, relativist?
>>>
>>> FUNDAMENTALS!
>
> What is the principle of least action in classical mechanics?
> In Classical Mechanics when a particle moves from one initial point to a
> final point the path that it will follow is the one
> where action is minimum.
> ------------
> The stationary-action principle – also known as the principle of least
> action – is a variational principle that, when applied to
> the action of a mechanical system, yields the equations of motion for that system. ...
> -----------
> What is classical mechanics action?
> More formally, action is a mathematical functional which takes the
> trajectory, also called path or history, of the system as its
> argument and has a real number as its result. ... Generally, the action
> takes different values for different paths.
> -----------
> https://www.feynmanlectures.caltech.edu/II_19.html
>
> The Principle of Least Action
>
> “Mr. Bader told me the following: Suppose you have a particle (in a
> gravitational field, for instance) which starts somewhere and
> moves to some other point by free motion—you throw it, and it goes up and
> comes down (Fig. 19–1). It goes from the original
> place to the final place in a certain amount of time. Now, you try a
> different motion. Suppose that to get from here to there, it went
> as shown in Fig. 19–2 but got there in just the same amount of time. Then
> he said this: If you calculate the kinetic energy at every
> moment on the path, take away the potential energy, and integrate it over
> the time during the whole path, you’ll find that the number
> you’ll get is bigger than that for the actual motion.
>
> “In other words, the laws of Newton could be stated not in the form F=ma
> but in the form: the average kinetic energy less the average potential
> energy is as little as possible for the path of an object going from one point to another.
>
> “Let me illustrate a little bit better what it means. If you take the
> case of the gravitational field, then if the particle has the path x(t)
> (let’s just take one dimension for a moment; we take a trajectory that
> goes up and down and not sideways), where x is the height above the
> ground, the kinetic energy is 1/2m(dx/dt)², and the potential energy at
> any time is mgx. Now I take the kinetic energy minus the potential energy
> at every moment along the path and integrate that with respect to time
> from the initial time to the final time. Let’s suppose that at the
> original time t1 we started at some height and at the end of the time t2
> we are definitely ending at some other place (Fig. 19–3).
>
> “Then the integral is ∫ [1/2m(dx/dt)²−mgx] dt, between t1 and t2
>
> The actual motion is some kind of a curve—it’s a parabola if we plot
> against the time—and gives a certain value for the integral. But we could
> imagine some other motion that went very high and came up and down in
> some peculiar way (Fig. 19–4). We can calculate the kinetic energy minus
> the potential energy and integrate for such a path … or for any other path we want.
>
> The miracle is that the true path is the one for which that integral is least.
> ---------
>
> Enough? I can find more trusted sources about the ORIGINAL PRINCIPLE OF
> LEAST ACTION (not the relativistic shit, staining everything!).
>

There! See? That wasn’t so hard was it? Such a good lad. Good for you. The
Feynman lectures was the first on my list to recommend if you had asked.
And if you had looks this up FIRST rather than yap-flapping for — what? — a
week and a half, then you would have needed a step ladder to crawl out of
the hole you kept digging.

And if you go on in the lectures, he’ll actually give some worked examples
of using the Euler Lagrange equations to find equations of motion.

There are even YouTube videos, including one I watched just about 2 weeks
ago, using the Lagrangian to solve the motion of a bead sliding
frictionlessly on a parabolic wire, where the wire is accelerated
horizontally. I’m sure you can find that too. Might take some time away
from yapping to just watch it though.

--
Odd Bodkin — Maker of fine toys, tools, tables

On Tuesday, January 18, 2022 at 7:59:47 PM UTC-3, Paul B. Andersen wrote:
> Den 18.01.2022 21:41, skrev Richard Hertz:
> > On Tuesday, January 18, 2022 at 9:25:29 AM UTC-3, Paul B. Andersen wrote:
> >> Den 17.01.2022 20:36, skrev Richard Hertz:
> <unsnip>>>> Fucking ignorant you all!
> >>>
> >>> Lagrangians, Hamiltonian and similar shit are basically Newton's laws for mechanical energy, WHICH DISIPATES with enough time (say a
> >>> couple of billion years for celestial mechanics).
> >>>
> >>> That's why dL/dt = 0 and its integral renders L = CONSTANT.
> I can understand why you snipped your own wrong statement! :-D
> >> I suppose that L is the Lagrangian.
> >>
> >> L = T - V kinetic energy MINUS potential energy.
> >>
> >> L is constant only if T and V both are constant.
> >> (Example: planet in circular orbit)
> >>
> >> But in the general case, both T and V may change.
> >> (Example planet in orbit with eccentricity ≠ 0)
> >>
> >> Since the total energy E = T+V = constant,
> >> dT/dt = - dV/dt
> >>
> >> dL/dt = 2dT/dt
> Is the following supposed to be a defence for your claim that dL/dt = 0?
> >
> > Mr. Astrophysicist, shame on you!
> >
> > The most elementary application for a system with conserved energy is with celestial bodies orbitals.
> >
> > This equation is valid for bounded (planets and their moons) or unbounded orbits (asteroids that pass by).
> >
> > 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E
> Yes, but this is the total energy, not the Lagrangian.
>
> T = 1/2 m [r² (dθ/dt)² + (dr/dt)²]
> V = – GMm/r
>
> L = T-V = 1/2 m [r² (dθ/dt)² + (dr/dt)²] + GMm/r
>
> E is constant, L is NOT.
>
> The next step is to use the Euler-Lagrange equation
> which will in this case give two equations:
>
> d/dt(∂L/∂ṙ) = ∂L/∂r
>
> d/dt(∂L/∂θ̇ ) = ∂L/∂θ
> >
> > For bounded orbits, E < 0 and 0 <= e < 1 (excentricity of the trajectory; i.e.: e = 0.2025 for Mercury).
> >
> > In the case of unbounded hyperbolic orbits, E > 0 and e > 1.
> >
> > So, at infinity, where gravitational potential goes to ZERO, and angular moment vanished because θ reached a limit value,
> > so dθ/dt = 0, only remains KE at infinity: E = 1/2 m v²∞.
> But still: T + V = E = const
>
> lim T = const when r -> ∞
> lim V = 0 when r -> ∞
> lim E = const when r -> ∞ as always
> >
> > Your dT/dt = 0 at infinity because T = constant = E
> >
> > but your dV/dt --> 0 just because V ---> 0 for distances very far away from the gravitational source (the Sun, in this case).
> So?
>
> dV/dt -> 0 and dT/dt -> 0 when r -> ∞
>
>
> What was your point?
> >
> > This is the fate or a one time flyby of an asteroid/comet if E > 0 (hyperbolic trajectory).
> Quite. But T+V = E is still constant
> >
> > In some cases, E = 0 and e = 1. It describes a parabolic trajectory, an exceptional case.
> Quite. But T+V = E is still constant
> >
> > Your dT/dt = - dV/dt applies for elliptic orbits only.
> No, it follows trivially from E = T + V. Always.
>
> dT/dt + dv/dt = dE/dt = 0
> dT/dt = -dV/dt
> >
> > and your dL/dt = 2dT/dt IS WRONG! We said that energy is CONSERVED, as well as angular moment.
> L = T - V
> dL/dt = dT/dt - dV/dt = 2dT/dt
>
> BECAUSE total energy is CONSERVED
>
> >
> > Go to the corner!
> >
>
> No, I will leave you there alone.
>
> --
> Paul
>
> https://paulba.no/

E = T + V
L = T - V
E + L = 2T
d/dt(E + L) = 2 dT/dt
dL/dt = 2 dT/dt

I concede, but I had fun and you went nuts. LOL.

On Tuesday, January 18, 2022 at 9:44:22 PM UTC-3, Richard Hertz wrote:
> On Tuesday, January 18, 2022 at 7:59:47 PM UTC-3, Paul B. Andersen wrote:
> > Den 18.01.2022 21:41, skrev Richard Hertz:
> > > On Tuesday, January 18, 2022 at 9:25:29 AM UTC-3, Paul B. Andersen wrote:
> > >> Den 17.01.2022 20:36, skrev Richard Hertz:
> > <unsnip>>>> Fucking ignorant you all!
> > >>>
> > >>> Lagrangians, Hamiltonian and similar shit are basically Newton's laws for mechanical energy, WHICH DISIPATES with enough time (say a
> > >>> couple of billion years for celestial mechanics).
> > >>>
> > >>> That's why dL/dt = 0 and its integral renders L = CONSTANT.
> > I can understand why you snipped your own wrong statement! :-D
> > >> I suppose that L is the Lagrangian.
> > >>
> > >> L = T - V kinetic energy MINUS potential energy.
> > >>
> > >> L is constant only if T and V both are constant.
> > >> (Example: planet in circular orbit)
> > >>
> > >> But in the general case, both T and V may change.
> > >> (Example planet in orbit with eccentricity ≠ 0)
> > >>
> > >> Since the total energy E = T+V = constant,
> > >> dT/dt = - dV/dt
> > >>
> > >> dL/dt = 2dT/dt
> > Is the following supposed to be a defence for your claim that dL/dt = 0?
> > >
> > > Mr. Astrophysicist, shame on you!
> > >
> > > The most elementary application for a system with conserved energy is with celestial bodies orbitals.
> > >
> > > This equation is valid for bounded (planets and their moons) or unbounded orbits (asteroids that pass by).
> > >
> > > 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E
> > Yes, but this is the total energy, not the Lagrangian.
> >
> > T = 1/2 m [r² (dθ/dt)² + (dr/dt)²]
> > V = – GMm/r
> >
> > L = T-V = 1/2 m [r² (dθ/dt)² + (dr/dt)²] + GMm/r
> >
> > E is constant, L is NOT.
> >
> > The next step is to use the Euler-Lagrange equation
> > which will in this case give two equations:
> >
> > d/dt(∂L/∂ṙ) = ∂L/∂r
> >
> > d/dt(∂L/∂θ̇ ) = ∂L/∂θ
> > >
> > > For bounded orbits, E < 0 and 0 <= e < 1 (excentricity of the trajectory; i.e.: e = 0.2025 for Mercury).
> > >
> > > In the case of unbounded hyperbolic orbits, E > 0 and e > 1.
> > >
> > > So, at infinity, where gravitational potential goes to ZERO, and angular moment vanished because θ reached a limit value,
> > > so dθ/dt = 0, only remains KE at infinity: E = 1/2 m v²∞.
> > But still: T + V = E = const
> >
> > lim T = const when r -> ∞
> > lim V = 0 when r -> ∞
> > lim E = const when r -> ∞ as always
> > >
> > > Your dT/dt = 0 at infinity because T = constant = E
> > >
> > > but your dV/dt --> 0 just because V ---> 0 for distances very far away from the gravitational source (the Sun, in this case).
> > So?
> >
> > dV/dt -> 0 and dT/dt -> 0 when r -> ∞
> >
> >
> > What was your point?
> > >
> > > This is the fate or a one time flyby of an asteroid/comet if E > 0 (hyperbolic trajectory).
> > Quite. But T+V = E is still constant
> > >
> > > In some cases, E = 0 and e = 1. It describes a parabolic trajectory, an exceptional case.
> > Quite. But T+V = E is still constant
> > >
> > > Your dT/dt = - dV/dt applies for elliptic orbits only.
> > No, it follows trivially from E = T + V. Always.
> >
> > dT/dt + dv/dt = dE/dt = 0
> > dT/dt = -dV/dt
> > >
> > > and your dL/dt = 2dT/dt IS WRONG! We said that energy is CONSERVED, as well as angular moment.
> > L = T - V
> > dL/dt = dT/dt - dV/dt = 2dT/dt
> >
> > BECAUSE total energy is CONSERVED
> >
> > >
> > > Go to the corner!
> > >
> >
> > No, I will leave you there alone.
> >
> > --
> > Paul
> >
> > https://paulba.no/
> E = T + V
> L = T - V
> E + L = 2T
> d/dt(E + L) = 2 dT/dt
> dL/dt = 2 dT/dt
>
> I concede, but I had fun and you went nuts. LOL.

But, on a second thought, dT/dt = 0 on hyperbolic trajectories far away from the Sun.

So I retire my concession and stand with dL/dt = 0 in celestial mechanics, not a fucking spring.

On Tuesday, January 18, 2022 at 5:57:15 AM UTC-8, bodk...@gmail.com wrote:
> Richard Hertz <hert...@gmail.com> wrote:
> > On Monday, January 17, 2022 at 9:15:15 PM UTC-3, Ross A. Finlayson wrote:
> >
> > <snip>
> >
> >> You mean d/dt dV - dT = 0 ?
> >
> > No, I mean that given E = T - V = 1/2 ∑ᴺ mᵢ vᵢ² - V(r⃗₁, r⃗₂, r⃗₃, r⃗₄,
> > ..., r⃗ᶰ) , for conservative systems like in newtonian gravity.
> E = T + U, not T - U.
>
> Perhaps you are confusing the Lagrangian with the total energy.
> >
> > This implies to solve a Lagrangian expressed as:
> >
> > ∂L/∂r⃗ᵢ - d/dt ∂L/∂ṙ⃗ᵢ + ∑ᴱ λᵦ ∂fᵦ/∂r⃗ᵤ = 0 , where i: 1, 2, 3,
> > ...., N particles and u: 1, 2, 3, ..., E constraints.
> Conservation of energy implies no such thing. The principle of stationary
> action implies the Euler-Lagrange equations you just learned.
> >
> > ∂r⃗ᵢ ≡ (∂/∂xᵢ , ∂/∂yᵢ , ∂/∂zᵢ). Use the same notation for ∂ṙ⃗ᵢ .
> >
> > At any case, ask Bodkin, who knows this shit in depth.
> Well, I know it more than you do. But if you scramble fast enough, you
> might learn something about it in a hurry.
> >
> > Also, ask him how the first order Lagrangian derives in this simple
> > expression used for planetary motion (even Einstein used it):
> >
> > 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E
> >
> > or
> >
> > d²r/dt² + μ/r² = 0 ------> 1/2 v² - μ/r = constant = E (all derived
> > from Newton-Euler m d²r⃗/dt² = - GMmr⃗/r³).
> >
> >
> >

Thanks, digesting - that's part of how energy is attached to force.

d/dt dV - dT = 0

dV - dT = 0t

"... the smooth path x_0 is a stationary point of S if all its directional
derivatives at x_0 vanish, ...".

"The Lagrangian is a function of time since the Lagrangian density
has implicit space dependence via the fields, and may have explicit
spatial dependence, but these are removed in the integral, leaving
only time in as the variable for the Lagrangian."

About representing energy and average energy, and normalization,
yes I am talking about what happens after the Lagrangian and
total energy are both zero or same, _and_, about the energy
computed from rest and the energy computed from motion,
as having effectively different Lagrangians, because the one's
integral is the others' derivative.

Ross A. Finlayson wrote:
> On Tuesday, January 18, 2022 at 5:57:15 AM UTC-8, bodk...@gmail.com wrote:
>> Richard Hertz <hert...@gmail.com> wrote:
>>> On Monday, January 17, 2022 at 9:15:15 PM UTC-3, Ross A. Finlayson wrote:
>>>
>>> <snip>
>>>
>>>> You mean d/dt dV - dT = 0 ?
>>>
>>> No, I mean that given E = T - V = 1/2 ∑ᴺ mᵢ vᵢ² - V(r⃗₁, r⃗₂, r⃗₃, r⃗₄,
>>> ..., r⃗ᶰ) , for conservative systems like in newtonian gravity.
>> E = T + U, not T - U.
>>
>> Perhaps you are confusing the Lagrangian with the total energy.
>>>
>>> This implies to solve a Lagrangian expressed as:
>>>
>>> ∂L/∂r⃗ᵢ - d/dt ∂L/∂ṙ⃗ᵢ + ∑ᴱ λᵦ ∂fᵦ/∂r⃗ᵤ = 0 , where i: 1, 2, 3,
>>> ...., N particles and u: 1, 2, 3, ..., E constraints.
>> Conservation of energy implies no such thing. The principle of stationary
>> action implies the Euler-Lagrange equations you just learned.
>>>
>>> ∂r⃗ᵢ ≡ (∂/∂xᵢ , ∂/∂yᵢ , ∂/∂zᵢ). Use the same notation for ∂ṙ⃗ᵢ .
>>>
>>> At any case, ask Bodkin, who knows this shit in depth.
>> Well, I know it more than you do. But if you scramble fast enough, you
>> might learn something about it in a hurry.
>>>
>>> Also, ask him how the first order Lagrangian derives in this simple
>>> expression used for planetary motion (even Einstein used it):
>>>
>>> 1/2 m [r² (dθ/dt)² + (dr/dt)²] – GMm/r = E
>>>
>>> or
>>>
>>> d²r/dt² + μ/r² = 0 ------> 1/2 v² - μ/r = constant = E (all derived
>>> from Newton-Euler m d²r⃗/dt² = - GMmr⃗/r³).
>>>
>>>
>>>
>
> Thanks, digesting - that's part of how energy is attached to force.
>
> d/dt dV - dT = 0
>
> dV - dT = 0t
>
> "... the smooth path x_0 is a stationary point of S if all its directional
> derivatives at x_0 vanish, ...".
>
> "The Lagrangian is a function of time since the Lagrangian density
> has implicit space dependence via the fields, and may have explicit
> spatial dependence, but these are removed in the integral, leaving
> only time in as the variable for the Lagrangian."
>
> About representing energy and average energy, and normalization,
> yes I am talking about what happens after the Lagrangian and
> total energy are both zero or same, _and_, about the energy
> computed from rest and the energy computed from motion,
> as having effectively different Lagrangians, because the one's
> integral is the others' derivative.

Ross, you're using a random text generator fed by a few articles,
right? Why that? What's the point?

Den 19.01.2022 01:52, skrev Richard Hertz:
> On Tuesday, January 18, 2022 at 9:44:22 PM UTC-3, Richard Hertz wrote:

>>>> On Tuesday, January 18, 2022 at 9:25:29 AM UTC-3, Paul B. Andersen wrote:
>>>>> Den 17.01.2022 20:36, skrev Richard Hertz:
>>>>>>
>>>>>> Lagrangians, Hamiltonian and similar shit are basically Newton's laws for mechanical energy, WHICH DISIPATES with enough time (say a
>>>>>> couple of billion years for celestial mechanics).
>>>>>>
>>>>>> That's why dL/dt = 0 and its integral renders L = CONSTANT.

>>>>> I suppose that L is the Lagrangian.
>>>>>
>>>>> L = T - V kinetic energy MINUS potential energy.
>>>>>
>>>>> L is constant only if T and V both are constant.
>>>>> (Example: planet in circular orbit)
>>>>>
>>>>> But in the general case, both T and V may change.
>>>>> (Example planet in orbit with eccentricity ≠ 0)
>>>>>
>>>>> Since the total energy E = T+V = constant,
>>>>> dT/dt = - dV/dt
>>>>>
>>>>> dL/dt = 2dT/dt

>> E = T + V
>> L = T - V
>> E + L = 2T
>> d/dt(E + L) = 2 dT/dt
>> dL/dt = 2 dT/dt
>>
>> I concede, but I had fun and you went nuts. LOL.

The glimpse of sensibleness lasted only for eight minutes:

>
> But, on a second thought, dT/dt = 0 on hyperbolic trajectories far away from the Sun.

So if dT/dt -> 0 when r -> ∞ ,
then L is always constant in celestial mechanics?

>
> So I retire my concession and stand with dL/dt = 0 in celestial mechanics, not a fucking spring.
>

Still having fun? :-D

--
Paul

https://paulba.no/

On Wednesday, January 19, 2022 at 5:01:28 AM UTC-3, Paul B. Andersen wrote:

<snip>

> >>>>>> Lagrangians, Hamiltonian and similar shit are basically Newton's laws for mechanical energy, WHICH DISIPATES
> >>>>>> with enough time (say acouple of billion years for celestial mechanics).
> >>>>>>
> >>>>>> That's why dL/dt = 0 and its integral renders L = CONSTANT.

THIS IS AN ASSERTION OF MY OWN, DERIVED FROM THE FACT THAT THE SOLAR SYSTEM IS KNOWN AS BORN 4.5 BILLION YEARS AGO.
It's not required to be a genius to understand that in such timespan, the stability has been transitory all the time. Only in lapses of, maybe,
10 million years, it could resemble some sort of stability. There are 450 time lapses of this amount in the life of the SS.

When I wrote that any system of equations should incorporate instability with time, I'm referring to the above. But this is the field of
cosmologist, I believe, more than of astrophysicists.

It's notorious that, in celestial mechanics, instantaneous time is not incorporated. Only periods in orbits. Yet, for eccentric trajectories,
scientists have managed to incorporate it somehow, like when calculating final angles and velocities in the analysis of asteroids
and comets.

MAYBE, it's due that newtonian mechanics is based on instantaneous actions at a distance, which prevent to represent the path of
any free body described by u(x, y, z, t) in "real time".

But that the Solar System is not stable is for me a certainty, based on common sense (which Bodkin depict). Just give it enough time.

> >>>>> I suppose that L is the Lagrangian.
> >>>>>
> >>>>> L = T - V kinetic energy MINUS potential energy.
> >>>>>
> >>>>> L is constant only if T and V both are constant. (Example: planet in circular orbit)
> >>>>>
> >>>>> But in the general case, both T and V may change. (Example planet in orbit with eccentricity ≠ 0)
> >>>>>
> >>>>> Since the total energy E = T+V = constant, dT/dt = - dV/dt
> >>>>>
> >>>>> dL/dt = 2dT/dt

> >> E = T + V
> >> L = T - V
> >> E + L = 2T
> >> d/dt(E + L) = 2 dT/dt
> >> dL/dt = 2 dT/dt
> >>
> >> I concede, but I had fun and you went nuts. LOL.

> The glimpse of sensibleness lasted only for eight minutes:
> >
> > But, on a second thought, dT/dt = 0 on hyperbolic trajectories far away from the Sun.

REGARDING THESE THINGS, I'VE MADE PUBLIC HERE THAT I'M LEARNING THIS TOPIC SINCE I GOT INVOLVED WITH
THE PROBLEM OF MERCURY AND EINSTEIN, MONTHS AGO. LATER CAME LIGHT DEFLECTION.

NOW I'M STUDYING HYPERBOLIC TRAJECTORIES WITHIN SOLAR SYSTEMS WITH THE BEST AVAILABLE MATERIAL.

I ALSO WROTE THAT I LEARN WITH A REFINEMENT LOOP. WITH EACH LOOP, I KNOW MORE.

For the rest, enervating Bodkin is priceless. Sorry that you meddle.

> So if dT/dt -> 0 when r -> ∞ ,
> then L is always constant in celestial mechanics?
> >
> > So I retire my concession and stand with dL/dt = 0 in celestial mechanics, not a fucking spring.
> >
> Still having fun? :-D
You bet, Paul. This a game for me. I'm not Brahe, Kepler or Euler. This is just a hobby.

Richard Hertz <hertz778@gmail.com> wrote:
> On Wednesday, January 19, 2022 at 5:01:28 AM UTC-3, Paul B. Andersen wrote:
>
> <snip>
>
>>>>>>>> Lagrangians, Hamiltonian and similar shit are basically Newton's
>>>>>>>> laws for mechanical energy, WHICH DISIPATES
>>>>>>>> with enough time (say acouple of billion years for celestial mechanics).
>>>>>>>>
>>>>>>>> That's why dL/dt = 0 and its integral renders L = CONSTANT.
>
> THIS IS AN ASSERTION OF MY OWN, DERIVED FROM THE FACT THAT THE SOLAR
> SYSTEM IS KNOWN AS BORN 4.5 BILLION YEARS AGO.

And given your history with statements you assert on your own, it’s very
likely worthless. Rest unread.

> It's not required to be a genius to understand that in such timespan, the
> stability has been transitory all the time. Only in lapses of, maybe,
> 10 million years, it could resemble some sort of stability. There are 450
> time lapses of this amount in the life of the SS.
>
> When I wrote that any system of equations should incorporate instability
> with time, I'm referring to the above. But this is the field of
> cosmologist, I believe, more than of astrophysicists.
>
> It's notorious that, in celestial mechanics, instantaneous time is not
> incorporated. Only periods in orbits. Yet, for eccentric trajectories,
> scientists have managed to incorporate it somehow, like when calculating
> final angles and velocities in the analysis of asteroids
> and comets.
>
> MAYBE, it's due that newtonian mechanics is based on instantaneous
> actions at a distance, which prevent to represent the path of
> any free body described by u(x, y, z, t) in "real time".
>
> But that the Solar System is not stable is for me a certainty, based on
> common sense (which Bodkin depict). Just give it enough time.
>
>
>
>>>>>>> I suppose that L is the Lagrangian.
>>>>>>>
>>>>>>> L = T - V kinetic energy MINUS potential energy.
>>>>>>>
>>>>>>> L is constant only if T and V both are constant. (Example: planet in circular orbit)
>>>>>>>
>>>>>>> But in the general case, both T and V may change. (Example planet
>>>>>>> in orbit with eccentricity ≠ 0)
>>>>>>>
>>>>>>> Since the total energy E = T+V = constant, dT/dt = - dV/dt
>>>>>>>
>>>>>>> dL/dt = 2dT/dt
>
>
>>>> E = T + V
>>>> L = T - V
>>>> E + L = 2T
>>>> d/dt(E + L) = 2 dT/dt
>>>> dL/dt = 2 dT/dt
>>>>
>>>> I concede, but I had fun and you went nuts. LOL.
>
>
>> The glimpse of sensibleness lasted only for eight minutes:
>>>
>>> But, on a second thought, dT/dt = 0 on hyperbolic trajectories far away from the Sun.
>
> REGARDING THESE THINGS, I'VE MADE PUBLIC HERE THAT I'M LEARNING THIS
> TOPIC SINCE I GOT INVOLVED WITH
> THE PROBLEM OF MERCURY AND EINSTEIN, MONTHS AGO. LATER CAME LIGHT DEFLECTION.
>
> NOW I'M STUDYING HYPERBOLIC TRAJECTORIES WITHIN SOLAR SYSTEMS WITH THE
> BEST AVAILABLE MATERIAL.
>
> I ALSO WROTE THAT I LEARN WITH A REFINEMENT LOOP. WITH EACH LOOP, I KNOW MORE.
>
> For the rest, enervating Bodkin is priceless. Sorry that you meddle.
>
>> So if dT/dt -> 0 when r -> ∞ ,
>> then L is always constant in celestial mechanics?
>>>
>>> So I retire my concession and stand with dL/dt = 0 in celestial
>>> mechanics, not a fucking spring.
>>>
>> Still having fun? :-D
>
> You bet, Paul. This a game for me. I'm not Brahe, Kepler or Euler. This is just a hobby.
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On Wednesday, January 19, 2022 at 2:08:04 PM UTC-3, bodk...@gmail.com wrote:

<snip>

> > THIS IS AN ASSERTION OF MY OWN, DERIVED FROM THE FACT THAT THE SOLAR
> > SYSTEM IS KNOWN AS BORN 4.5 BILLION YEARS AGO.

> And given your history with statements you assert on your own, it’s very likely worthless. Rest unread.

You shouldn't hate or despise my ideas because you have issues with me, Bodkin.

You're being judgmental, which goes against most of your written thoughts at this forum.

You know that Solar System was born billion of years ago, and reached this apparent stability not so much time ago,
in cosmological scales. That's why I divided the alleged age of 4.5 billion years in 450 lapses of 10 million years ago.

I could bet that the evolution and convergence to something similar to the current state consumed more than 90% of that time.

So, asymptotically in cosmological time, our current frame of time (N = 450) LOOKS LIKE STABLE, but actually it's not.

And this is ONE OF THE REASONS by which I wrote about my disbelief about conservation of energy and momentum IN THE LONG
TERM!

Anyone, with a functional brain and a bunch of neurons making synapses can SEE and UNDERSTAND such scenario.

Just play this movie IN YOUR MIND, starting with a laplacian nebulae, and roll on the next 449 frames, 10 million years each.

But (no pun intended) maybe you can't, because your brain didn't developed like mine, which is/was determined as highly visual
with 3D capabilities (colors included, plus motion). That's how I see things (WHATEVER) even before touching the first equation.

If this is your case, and I insist, I'm NOT SHOWING OFF, maybe it can explain many of our different points of view.

For me is crystal clear, before messing with math. The same thing happens with my vision of galaxies (many), their rotations, and
I can zoom in or zoom out.

I was born this way, which is the source of many controversies in the past with colleagues. Actually, it doesn't matter the discipline.

But don't hate me for having this "ability"?, which is more a source of trouble than a bridge, when I relate to others.